Problem: Simplify and expand the following expression: $ \dfrac{1}{a + 1}+ \dfrac{5}{a + 8}+ \dfrac{4}{a^2 + 9a + 8} $
First find a common denominator by finding the least common multiple of the denominators. Try factoring the denominators. We can factor the quadratic in the third term: $ \dfrac{4}{a^2 + 9a + 8} = \dfrac{4}{(a + 1)(a + 8)}$ Now we have: $ \dfrac{1}{a + 1}+ \dfrac{5}{a + 8}+ \dfrac{4}{(a + 1)(a + 8)} $ The least common multiple of the denominators is: $ (a + 1)(a + 8)$ In order to get the first term over $(a + 1)(a + 8)$ , multiply by $\dfrac{a + 8}{a + 8}$ $ \dfrac{1}{a + 1} \times \dfrac{a + 8}{a + 8} = \dfrac{a + 8}{(a + 1)(a + 8)} $ In order to get the second term over $(a + 1)(a + 8)$ , multiply by $\dfrac{a + 1}{a + 1}$ $ \dfrac{5}{a + 8} \times \dfrac{a + 1}{a + 1} = \dfrac{5(a + 1)}{(a + 1)(a + 8)} $ Now we have: $ \dfrac{a + 8}{(a + 1)(a + 8)} + \dfrac{5(a + 1)}{(a + 1)(a + 8)} + \dfrac{4}{(a + 1)(a + 8)} $ $ = \dfrac{ a + 8 + 5(a + 1) + 4} {(a + 1)(a + 8)} $ Expand: $ = \dfrac{a + 8 + 5a + 5 + 4}{a^2 + 9a + 8} $ $ = \dfrac{6a + 17}{a^2 + 9a + 8}$